Zero set of a Wiener Process Consier $B(t)$ a standard Wiener Process on $[0,1]$ and let $$  \gamma = \sup \{t \in [0,1] : B(t) = 0     \}  $$
I want to show that almost surely there exists sequences $(t_n)_{n=1}^{\infty}$, $(s_n)_{n=1}^{\infty}$ such that $t_n < s_n < \gamma $ with $t_n \uparrow \gamma$ such that the following holds $$ B(t_n) < 0 \; \; \; \; , \; \; \; B(s_n) > 0  \; \; \; \; \; \forall n  $$
I was thinking of using the fact that almost surely the set is not finite and so there exists a sequence $(p_n)_{n=1}^{\infty}$ converging to $\gamma$ with $B(p_n) = 0 $ but then I'm not sure how to show that the process would almost surely change signs between each of the $p_n$.
Is there a better way to proceed?
 A: Recall that the time-inverse
$$W_t := t B_{1/t}, \qquad W_0 := 0$$
defines a Brownian motion. By definition, we have
$$\begin{align*} \gamma &= \sup\{t \in [0,1]; B_t = 0\} = \inf\{t \geq 1; W_t=0\}.\end{align*}$$
Since $\gamma$ defines a stopping time (with respect to the natural filtration of $(W_t)_{t \geq 0}$) which is almost surely finite, the process
$$X_t := W_{\gamma+t}-W_{\gamma}, \qquad t \geq 0,$$
is again a Brownian motion.
Denote by $A$ the set of $\omega \in \Omega$ such that there exist sequences $(s_n)_{n \in \mathbb{N}}$ and $(t_n)_{n \in \mathbb{N}}$ such that $t_n<s_n<\gamma(\omega)$, $t_n \uparrow \gamma(\omega)$ and $B(t_n,\omega)<0$, $B(s_n,\omega)>0$ for all $n \in \mathbb{N}$. Then
$$A^c \subseteq \bigcup_{n \in \mathbb{N}} \left(\left\{ \sup_{t \leq 1/n} X_t = 0 \right\} \cup \left\{ \inf_{t \leq 1/n} X_t = 0 \right\} \right).$$
Consequently, we are done if we can show the following statement.

Let $(X_t)_{t \geq 0}$ be a Brownian motion. Then $$\mathbb{P} \left( \sup_{t \leq T} X_t = 0 \right) = \mathbb{P} \left( \inf_{t \leq T} X_t = 0 \right)=0$$ for any $T>0$.

Proof: By the reflection principle, we know that $M_T := \sup_{t \leq T} X_t$ has the same distribution as $|X_T|$. Therefore,
$$\mathbb{P}(M_T=0) = \mathbb{P}(|X_T|=0)=0.$$
Similarly, by the reflection principle, $m_T := \inf_{t \leq T} X_t$ equals in distribution $-|X_T|$, and again this implies
$$\mathbb{P}(m_T=0) = \mathbb{P}(-|X_T|=0)=0.$$
